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Evaluate  \mathrm{\lim _{x \rightarrow 0} \frac{\tan (\tan x)-\sin (\sin x)}{\tan x-\sin x}}

Option: 1

1


Option: 2

0


Option: 3

-1


Option: 4

2


Answers (1)

best_answer

Then, by mean value theorem, there is \mathrm{c \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)} such that
\mathrm{\frac{\tan (\tan x)-\tan (\sin x)}{\tan x-\sin x}=\sec ^{2} c}

and between \mathrm{\sin x} and \mathrm{\tan x}. Then \mathrm{\lim _{x \rightarrow 0} \frac{\tan (\tan x)-\tan (\sin x)}{\tan x-\sin x}=1}. Next, we will show that \mathrm{\lim _{x \rightarrow 0} \frac{\tan (\sin x)(1-\cos (\sin x))}{\tan x-\sin x}=1}.

\mathrm{\lim _{x \rightarrow 0} \frac{\tan (\sin x)(1-\cos (\sin x))}{\tan x-\sin x} =\lim _{x \rightarrow 0} \frac{\tan (\sin x)(1-\cos (\sin x)) \cos x}{\sin x(1-\cos x)} }
                                                                \mathrm{=\lim _{x \rightarrow 0} \frac{\tan (\sin x)}{\sin x} \cdot \frac{1-\cos (\sin x)}{\sin ^{2} x} \cdot \frac{\sin ^{2} x}{1-\cos x} \cdot \cos x}
                                                                \mathrm{ =\lim _{x \rightarrow 0} \frac{\tan (\sin x)}{\sin x} \lim _{x \rightarrow 0} \frac{1-\cos (\sin x)}{\sin ^{2} x} \lim _{x \rightarrow 0}(1+\cos x) \lim _{x \rightarrow 0} \cos x \\ }
                                                                 \mathrm{ =1 \cdot \frac{1}{2} \cdot 2 \cdot 1 }
                                                                  \mathrm{ =1 }.
         

Posted by

SANGALDEEP SINGH

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